Evaluate $\int x\sec x\tan x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $ x\sec x+C$ (Choice B) B $ \tan x-\ln|\sec x+\tan x|+C$ (Choice C) C $ x\sec x-\ln|\sec x+\tan x|+C$ (Choice D) D $ \ln|\sec x+\tan x|+C$
Solution: This is an integration by parts problem, since the derivative of $x$ is constant and the antiderivative of $\sec x\,\tan x$ is easy. Let $u = x$ and $dv = \sec x\tan x\,dx$. Then $du=dx$ and $v=\sec x$. We now get $\int x\sec x\tan x\,dx= x\sec x-\int\sec x\,dx$ Finishing the integration, we have $\begin{aligned} &\phantom{=}\int x\sec x\tan x\,dx \\\\ &= x\sec x-\int\sec x\,dx \\\\ &=x\sec x-\ln|\sec x+\tan x|+C \end{aligned}$